Predictive Control- Exercise Session 4 Adaptive Control: Self Tuning Regulators and Model Reference Adaptive Systems

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1 Predictive Control- Exercise Session 4 Adaptive Control: Self Tning Reglators and Model Reference Adaptive Systems 1. Indirect Self Tning Reglator: Consider the system where G(s)=G 1 (s)g 2 (s) G 1 (s)= a s+b G 2 (s)= c s+d Hereaandbarenknownparametersandcanddareknown.Thiscold for example represent a system where the plant is known bt where certain sensordynamicsarenknown.thesystemistobecontrolledinschaway thattheclosedloopsystemisgivenby: G m (s)= ω 2 s 2 +2ωζs+ ω 2 a. Constrct a discrete time indirect self tning reglator withot zero cancellation. b. Constrct a discrete time indirect self tning reglator with zero cancellation. Soltion Webeginbysamplingboththeplantandthedesiredclosedloop: H(z)= b 0z+b 1 z 2 +a 1 z+a 2 H m (z)= b m0z+b m1 z 2 +a m1 z+a m2 Since the sampled plant s discrete time parameters may depend on both the known and nkown continos time parameters, we will proceed as thogh all parameters were nkown. Since we are designing an indirect STR, we reqire a linear regression model for the plant s parameters. This is given by: θ=(b o b 1 a 1 a 2 ) T φ(t)=((t 1) (t 2) y(t 1) y(t 2)) a. First we will design the controller withot cancelling the process zero(i.e. B + =1).UsingthecasalityconditiondegA c 2degA 1wefindthatA c msthaveatleastdegree3.sinceweknowthatthedegreeofais2,the degreeofrmstthereforebeatleast1.theminimmdegreedesignisachieved by choosing the smallest possible degrees of the design polynomials, :15 1

2 sowegetdega c =3anddegR=degS=1.TheDiophantineeqationfor this problem is: (z 2 +a 1 z+a 2 )(z+r 1 )+(b 0 z+b 1 )(s 0 z+s 1 )=(z 2 +a m1 z+a m2 )(z+a o1 ) Identification of coefficients of eqal powers of z gives: z 2 :a 1 +r 1 +b 0 s 0 =a m1 +a o1 z 1 :a 2 +a 1 r 1 +b 1 s 0 +b 0 s 1 =a m1 a o1 +a m2 z 0 :a 2 r 1 +b 1 s 1 =a m2 a o1 The soltion to these linear eqations is: where: r 1 = b2 1 n 1 b 0 b 1 n 2 +a o1 a m2 b 2 0 b 2 0 a 2 a 1 b 0 b 1 +b 2 1 s 0 = n 1 r 1 b 0 s 1 = b 0n 2 b 1 n 1 r 1 (a 1 b 0 b 1 ) b 2 0 n 1 =a m1 +a o1 a 1 n 2 =a m1 a o1 +a m2 a 2 b. Thecontrollerwillnowbedesignedsingzerocancellation( B + = z+ b 1 /b 0 ).Theordersofthepolynomialsarechosentobethesameasabove, btnowafactorofb + maybecancelledfromthediophantineeqation.in this case the Diophantine eqation gives: (z 2 +a 1 z+a 2 )1+b 0 (s 0 z+s 1 )=z 2 +a m1 z+a m2 Identification of coefficients of eqal powers of z gives: The controller is ths given by: z 1 :a 1 +b 0 s 0 =a m1 s 0 = a m1 a 1 b 0 z 0 :a 2 +b 0 s 1 =a m2 s 1 = a m2 a 2 b 0 R(z)=z+b 1 /b 0 S(z)=s 0 z+s 1 T(z)=t 0 z where t 0 = 1+a m1+a m2 b 0 Toexaminethebehaviorofthesystemswehavedesigned,wemayse simlations. Simlation of the system withot zero cancellation is shown in Figre 1. Simlation reslts for the system desinged with cancellation areshowninfigre2.itcanbeseenthatthecontrolsignalinthedesign :15 2

3 c y Figr 1 Simlation of Problem 1a. Process otpt and control signal are shown for the indirect self-tning reglator when the process zero is not cancelled. with cancellation exhibits ringing. This is the reslt of cancelling a poorly dampedprocesszero.inthiscasetheprocessisgivenby: whichhasazeroinz= H(z)= (z+0.936) (z 1)(z 0.819) 2. Direct Self Tning Reglator: Using the same plant and specification asinproblem1,design: a. A direct self tning reglator withot zero cancellation. b. A direct self tning reglator with zero cancellation. Soltion To obtain a direct self-tning reglator we start with the Diophantine design eqation: AR+BS=A m A o B + Let the design eqation operate on y: B + A m A o y=ary+bsy=br+bsy ( ) ( ) B B y=r +S y A o A m A o A m }{{}}{{} f y f :15 3

4 c y Figr 2 Simlation of Problem 1b. Process otpt and control signal are shown for the indirect self-tning reglator when the process zero is cancelled. Here, A m isthedesiredcharacteristicpolynomialand A o istheobserver polynomial.wenowhavearegressionmodelinwhichtherandspolynomialsareparametersandthefilteredinptandotptsignals f andy f areregressors.thswemayestimaterands. TheTpolynomialisgivenby: T= t oa o B m B wheret o ischosentogivethecorrectsteadystategain. a. Wewillbeginbydesigningwithotcancellation.WethenhaveB + =1and B =b 0 q+b 1.FromtheanalysisisoftheindirectSTRweknowthata firstorderobserverisreqired,i.e.a 0 =q+a o1.wehaveasbefore: ( ) ( ) B B y=r +S (1) A o A m }{{} f y A o A m }{{} y f SinceB isnotknownwecannotcalclate f andy f.onepossibilityisto rewrite Eqation 1 as: ( ) ( ) 1 1 y=rb }{{ } +SB A o A m }{{} y A o A m R S andtoestimater ands assecondorderpolynomialsandtocancelthe commonfactorb fromtheestimatedpolynomials.thisisdifficltbecase there will not be an exact cancellation. Another possibility is to se some estimateofb.athirdpossibilityistotrytoestimateb RandB Sasa bilinear problem :15 4

5 b. IfthepolynomialBiscanceledwehaveB + =z+b 1 /b 0,B =b 0.From theanalysisoftheindirectstrweknowthatnoobserverisneededinthis caseandthatthecontrollerhasthestrctredegr=degs=1.hence: y(t)=r ( bo A m (t) ) +S ( bo A m y(t) Sinceb o isnotknownweincldeitinthepolynomialrandsandestimate it.thepolynomialrthenisnotmonic.wehave: ( ) ( ) 1 1 y(t)=(r 0 q+r 1 ) (t) +(s 0 q+s 1 ) y(t) A m A m ToobtainadirectSTRwethsestimate: θ=(r 0 r 1 s 0 s 1 ) T byrls.thecaser 0 =0mstbetakencareofseparately.FrthermoreT hastheformt(q)=t 0 qwhere: To get nit steady state gain choose: BT AR+BS = Bt oq B + = t oq b }{{} o A m A m B t o =A m (1) InFigres3 6weshowsimlationwhenthemodelinEqation1issed with: B =1 B =q B = q B = q The simlation reslts for the case when the process zero is cancelled are showninfigre7. Again we see that cancellation of the process zero gives a ringing control signal.however,itisalsonotedthatthechoiceofb (whichisnknown) is critical for performance in the case where no zero is cancelled. 3. Model Reference Adaptive Control sing MIT Rle: In this problem we consider a linear process with the transfer fnction kg(s), where G(s) is known and k is an nknown parameter. Find a feedforward controller thatgivesasystemwiththetransferfnctiong m (s)=k 0 G(s)wherek 0 is a given constant. Use the controller strctre = θ c whereisthecontrolsignaland c thecommandsignal.usethemit rletopdatetheparameter θ,anddrawablockdiagramofthereslting adaptive system. ) :15 5

6 c y Figr 3 Simlation in Problem 2a. Process otpt and control signal are shown for the directself-tningreglatorwhentheprocesszeroisnotcanceledandwhenb =1. Soltion With the controller strctre = θ c the transfer fnction from command signal to the otpt becomes θ kg(s). ThistransferfnctioniseqaltoG m (s)iftheparameter θischosenas θ= k 0 k WewillnowsetheMIT-rletoobtainamethodforadjstingtheparameter θwhenkisnotknown.theerroris e=y y m =kg(p)θ c k 0 G(p) c where c isthecommandsignal,y m themodelotpt,ytheprocessotpt, θ the adjstable parameter, and p = d/dt the differential operator. The sensitivity derivative is given by e θ =kg(p) c= k k 0 y m The MIT rle then gives the following adaptation law dθ dt = γ k k 0 y m e= γy m e (2) where γ = γ k/k 0 hasbeenintrodcedinsteadof γ.noticethatinorder tohavethecorrectsignof γ itisnecessarytoknowthesignofk.eqation 2givesthelawforadjstingtheparameter.Ablockdiagramofthesystem isshowninfigre :15 6

7 c y Figr 4 Simlation in Problem 2a. Process otpt and control signal are shown for the directself-tningreglatorwhentheprocesszeroisnotcanceledandwhenb =q. c y Figr 5 Simlation in Problem 2a. Process otpt and control signal are shown for thedirectself-tningreglatorwhentheprocesszeroisnotcanceledandwhen B = (q+0.4)/ :15 7

8 c y Figr 6 Simlation in Problem 2a. Process otpt and control signal are shown for thedirectself-tningreglatorwhentheprocesszeroisnotcanceledandwhen B = (q 0.4)/0.6. c y Figr 7 Simlation in Problem 2b. Process otpt and control signal are shown for the direct self-tning reglator when the process zero is canceled :15 8

9 c θ 0 Π 1 s y m Σ 1 y s e θ Figr8 BlockdiagramofanMRASforadjstmentofafeedforwardgainbasedonthe MIT rle :15 9